Ten Coin Flips, Four Heads

Date: 05/08/2001 at 18:21:06
From: Timi
Subject: If you flip a coin 10 times, what is the probability of
getting at least 4 heads?
Hi!
My professor and I disagree about solving this problem:
If you flip a coin ten times, what is the probability of getting at
least four heads?
The way I see it is that there are 2^10 possible outcomes, since
each separate toss has 2 outcomes: head or tail, and there are
10 tosses. So
10C4 (or C10,4?) / 2^10 = 20.5%
The way my professor deals with this problem is:
"Add the ninth row and then the tenth row, by adding the two
numbers above in Pascal's triangle :
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
and we can write our equation:
1[(.5)^10] [(.5)^0] + 10[(.5)^9] [(.5)^1] + 45 [(.5)^8]
[(.5)^2]+120[(.5)^7] [(.5)^3] +210[(.5)^6] [(.5)^4]+252[(.5)^5]
[(.5)^5]+210[(.5)^4] [(.5)^6]+ etc.
but I can stop here because it says AT LEAST 4 and that means
10,9,8..4 so I can stop.
Now a shortcut. Because our probabilities are .5 heads and .5 tails,
each term is [(.5)^.5] or [1/2^10] or 1/(2^10) or 1/1024 so I can
just add the coefficients I need and divide by 1024.
Adding 1 10 45 120 210 252 210 = 848 which we divide by
1024 = 82.8% "
It seems a little bit complex to me, and our answers are different,
too. Could you please let me know what you think about this problem?
Thank you!
Timi